\newproblem{lay:2_2_25}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 2.2.25}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Consider $A=\begin{pmatrix}a & b \\ c & d\end{pmatrix}$. Show that if $ad-bc=0$, then the equation $A\mathbf{x}=\mathbf{b}$ has more than one solution. Why does
	this imply that $A$ is not invertible?
}{
  % Solution
	Let us reduce the augmented matrix $\left(\begin{array}{cc}A|\mathbf{b}\end{array}\right)$.
	\begin{center}
		$\left(\begin{array}{rrr}a & b & b_1 \\ c & d & b_2\end{array}\right) \sim 
		 \left(\begin{array}{rrr}a & b & b_1 \\ 0 & ad-bc & ab_2-cb_1\end{array}\right)$
	\end{center}
	In fact if $ad-bc=0$, the matrix equation may have infinite solutions (if $ab_2-cb_1=0$) or no solution at all (if $ab_2-cb_1\neq 0$). This implies that $A$ is not invertible
	because if it were invertible for any $\mathbf{b}\in\mathbb{R}^2$ the equation $A\mathbf{x}=\mathbf{b}$ would have a single solution.
}
\useproblem{lay:2_2_25}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
